Download Lecture 7 - Voltage Regulator and Half Wave Rectifier

Recall-Lecture 6
• Diode AC equivalent circuit – small signal analysis
– During AC analysis the diode is equivalent to a resistor,
rd
VDQ
+
-
IDQ
DC equivalent
rd
id
AC equivalent
DC ANALYSIS
DIODE = MODEL 1
,2 OR 3
CALCULATE DC
CURRENT, ID
AC ANALYSIS
CALCULATE
rd
DIODE =
RESISTOR, rd
CALCULATE AC
CURRENT, id
• Zener effect and Zener diode
– When a Zener diode is reverse-biased, it acts at
the breakdown region, when it is forward
biased, it acts like a normal PN junction diode
• Avalanche Effect
– Gain kinetic energy – hit another atom –
produce electron and hole pair
Model 1
V = 0
Model 2
V
Model 3
V and rf
Load Line  ID vs VD
Forward Biased, DC Analysis
At 300K VT = 0.026 V
Reverse Biased
Group 5
Materials
CHAPTER 2
Semiconductor:
Group 4 eg.
Silicon and
Germanium
Bandgap
Energy
Must perform DC
Analysis first to get
DC diode current, ID
PN junction
N-type
Insulator
Conductor
Semiconductor
AC Analysis
Thermal equilibrium,
depletion region
P-type
Group 3
Calculate
rd = VT / ID
Extrinsic
photodiode
Intrinsic
Other
types of
diode
Solar cells
LED
Zener Diode
Chapter 3
Diode Circuits
Voltage Regulator
Voltage Regulator - Zener Diode
A voltage regulator supplies constant voltage to a load.

The breakdown voltage of a Zener
diode is nearly constant over a wide
range of reverse-bias currents.

This make the Zener diode useful in a
voltage regulator, or a constantvoltage reference circuit.
3. The remainder of VPS
drops across Ri
2. The load
resistor sees
a constant
voltage
regardless of
the current
1. The Zener diode holds the
voltage constant regardless of
the current
Example
A Zener diode is connected in a voltage regulator circuit. It is given that VPS = 20V, the
Zener voltage, VZ = 10V, Ri = 222  and PZ(max) = 400 mW.
a. Determine the values of IL, IZ and II if RL = 380 .
b. Determine the value of RL that will establish PZ(max) = 400 mW in the diode.
ANSWER:
Part (a)
IL = 26.3 mA
IZ = 18.7 mA
II = 45 mA
ANSWER:
Part (b)
PZ = IZ VZ
IZ = 40 mA
IL = 45 -40 = 5 mA
 RL = 2 k
For proper function the circuit must satisfied the following conditions.
1. The power dissipation in the Zener diode is less than the rated value
2. When the power supply is a minimum, VPS(min), there must be minimum
current in the Zener diode IZ(min), hence the load current is a maximum,
IL(max),
3. When the power supply is a maximum, VPS(max), the current in the diode is a
maximum, IZ(max), hence the load current is a minimum, IL(min)
AND
Or, we can write
Considering designing this circuit by substituting IZ(min) = 0.1 IZ(max),
now the last Equation becomes:
Maximum power dispassion in the Zener diode is
EXAMPLE 1
Consider voltage regulator is used to power
the cell phone at 2.5 V from the lithium ion
battery, which voltage may vary between 3
and 3.6 V. The current in the phone will vary 0
(off) to 100 mA(when talking). Calculate the
value of Ri and the Zener diode power
dissipation
Solution:
The stabilized voltage VL = 2.5 V, so the Zener diode voltage must be VZ = 2.5 V. The
maximum Zener diode current is
The maximum power dispassion in the Zener diode is
The value of the current limiting resistance is
Rectifier
Rectifier Circuits

A DC power supply is required to bias all electronic circuits.

A diode rectifier forms the first stage of a dc power supply.
Diagram of an Electronic Power Supply

Rectification is the process of converting an alternating (ac) voltage
into one that is limited to one polarity.

Rectification is classified as half-wave or full-wave rectifier.
Rectifier Parameters
Relationship between the number of turns of a
step-down transformer and the input/output
voltages
𝑣𝑃
𝑣𝑆
=
𝑁1
𝑁2
The peak inverse voltage (PIV) of the diode is the peak value of the voltage
that a diode can withstand when it is reversed biased
Duty Cycle: The fraction of the wave cycle over which the diode is
conducting.
Half Wave Rectifier
•
vs< V, diode off, open circuit, no
current flow, Vo = 0V
vs > V, diode conducts, current flows,
v = vs – V
•
o
V
Equation of VO and current when diode is conducting
𝑣𝑂 = 𝑖𝐷 𝑅 = 𝑣𝑆 − 𝑉𝛾
𝑣𝑆 − 𝑉𝛾
𝑖𝐷 =
𝑅
vs < V, diode off, open circuit, no current flow, vo = 0V
• vs > V, diode conducts, current flows and vo = vs – V
•
Consider a sine wave where
v
m
vs = v
m sin
t and
v
m
is the peak value
Notice that the
peak voltage of Vo
is lower
V
vs > V
Example
Consider the rectifier circuit in the figure below. Let R = 1 k, and the diode
has the properties of V = 0.6 V and rf = 20 .
Assume
i.
ii.
v = 10 sin t (V)
s
Determine the peak value of the diode current
Sketch vO versus time, t. Label the peak value of vO.
v
s
SOLUTION
vO
vs > V